which of the statements is true?
oa. as x approaches positive infinity, f(x) exceeds g(x) and h(x).
ob. as x approaches positive infinity, g(x) exceeds f(x) and h(x).
oc. as x approaches positive infinity, h(x) converges with g(x).
od. as x approaches positive infinity, h(x) exceeds f(x) and g(x).

Question

Accepted Answer

# Explanation:
## Step1: Analyze the growth rate of linear, exponential and logarithmic functions
- $f(x)$ is a linear function ($y = x - 1$). The general form of a linear function is $y=mx + b$ with a constant slope $m$. As $x\to+\infty$, $y = mx + b$ grows at a constant rate.
- $g(x)$ is an exponential - like function (assuming $g(x)=2^{x}$ based on its shape). The general form of an exponential function is $y = a\cdot b^{x}+c$ ($b> 1$). As $x\to+\infty$, $y=a\cdot b^{x}+c$ ($b > 1$) grows much faster than a linear function. The growth rate of $y=a\cdot b^{x}+c$ ($b>1$) is proportional to the function's current value.
- $h(x)$ is a logarithmic - like function (assuming $h(x)=\log_{2}(x + 3)$ based on its shape). The general form of a logarithmic function is $y=\log_{b}(x - h)+k$. As $x\to+\infty$, $y = \log_{b}(x - h)+k$ grows, but its growth rate $\frac{dy}{dx}=\frac{1}{(x - h)\ln b}$ approaches $0$.

## Step2: Compare the functions as $x\to+\infty$
- For a linear function $y_1=mx + b$ ($m>0$), an exponential function $y_2=a\cdot b^{x}+c$ ($b > 1,a>0$) and a logarithmic function $y_3=\log_{d}(x - k)+l$ ($d>1$):
    - $\lim_{x\to+\infty}\frac{y_1}{y_2}=\lim_{x\to+\infty}\frac{mx + b}{a\cdot b^{x}+c}$. Using L'Hopital's rule (for $\frac{\infty}{\infty}$ form, if we consider the ratio of the derivatives), $\lim_{x\to+\infty}\frac{m}{a\cdot b^{x}\ln b}=0$. So $y_2$ (exponential - type function $g(x)$) grows faster than $y_1$ (linear function $f(x)$).
    - $\lim_{x\to+\infty}\frac{y_3}{y_2}=\lim_{x\to+\infty}\frac{\log_{d}(x - k)+l}{a\cdot b^{x}+c}$. Since the derivative of $\log_{d}(x - k)+l$ is $\frac{1}{(x - k)\ln d}$ and the derivative of $a\cdot b^{x}+c$ is $a\cdot b^{x}\ln b$, $\lim_{x\to+\infty}\frac{\frac{1}{(x - k)\ln d}}{a\cdot b^{x}\ln b}=0$. So $y_2$ (exponential - type function $g(x)$) grows faster than $y_3$ (logarithmic - type function $h(x)$).

# Answer:
B. As $x$ approaches positive infinity, $g(x)$ exceeds $f(x)$ and $h(x)$

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Question

Explanation:

Step1: Analyze the growth rate of linear, exponential and logarithmic functions

Step2: Compare the functions as \(x\to+\infty\)

Answer: